Proof of properties of Rhombus:

**i)** Diagonals of a Rhombus are perpendicular to each other

**Proof:** Let us consider triangles $POQ$ and $QOR$. We know that the diagonals bisect each other, so $PO$ = $OR$

**Given:** $PQRS$ is a Rhombus

$PO$ = $OR$ (Diagonals bisect each other)

$QO$ = $QO$ (reflexive property of congruency)

$PQ$ = $QR$ (Sides of Rhombus are equal as per properties of Rhombus)

Therefore, Triangle $POQ$ Congruent Triangle $QOR$ (By $SSS$ Property)

It is known to us that corresponding angles in congruent triangles are congruent. We see that angles $POQ$ and $QOR$ are corresponding angles. Thus, angle $POQ$ congruent angle $QOR$.

$O$ lies on the straight line $PR$.

Hence,

Angle $POR$ + angle $QOR$ = $180$degrees (supplementary angles)

Angle $POR$ + Angle $POR$ = $180$degrees

$2\ \times$ Angle $POR$ = $180$ (Addition property)

Angle $POR$ = $90$ (Division property)

Angle $QOR$ = Angle $POR$ = $90$ degrees (As these two angles are congruent)

Therefore, the diagonals of a Rhombus are perpendicular to each other.

**ii)** Diagonals of a Rhombus bisect each other

**Proof:** Let us consider triangles $POQ$ and $ROS$

Angle $QPO$ = Angle $SRO$ (Alternate Interior Angles as $PQ$ parallel to $SR$)

Angle $PQO$ = Angle $RSO$ (Alternate Interior Angles as $PQ$ parallel to $SR$)

Angle $POQ$ = Angle $SOR$ (Vertically opposite angles)

Therefore triangle $POQ$ Congruent triangle $SOR$ ($ASA$ property of congruency)

Thus, $PO$ = $OR$ (By CPCT)

$QO$ = $OS$ (By CPCT)

Hence, diagonals of a Rhombus bisect each other